Dear All, I am trying to brush up my communication theory fundamentals. I have simple(perhaps foolish!) question on intutive understanding of CDMA. Shannon says that the capacity of a channel is directly proportional to bandwidth of signal. In other others for a given SNR if increasing B would give me more capacity. DSSS CDMA uses the same principle to spread the symbol over a large bandwidth by multiplying it with narrow time pulses(chips). Can we intutively say therefore, that when we de-modulate our spreaded signal by remultiplying it with the PN sequene and then averaging it over symbol period, we are in-effect using the statistical property that noise over a large period(symbol period) will cancel out itself. In other words the larger averaging interval we have(symbol period) the more we lower noise, since noise is inherently random and gaussian. Can DSSS-CDMA be looked at from this perspective? regs ashu
DSSS-CDMA query
Started by ●August 10, 2010
Reply by ●August 10, 20102010-08-10
On 08/10/2010 08:29 AM, ashu wrote:> Dear All, > I am trying to brush up my communication theory fundamentals. I have > simple(perhaps foolish!) question on intutive understanding of CDMA. > > Shannon says that the capacity of a channel is directly proportional > to bandwidth of signal. In other others for a given SNR if increasing > B would give me more capacity.Yes, but that's maybe not a fair comparison, because keeping SNR constant and increasing bandwidth generally means that you're increasing power. The power capability of your transmitter directly relates both to cost of acquisition and cost of operation, and increasing the bandwidth increases the opportunity cost for _someone_, if not for you. So holding SNR constant and increasing bandwidth is increasing the overall cost of your system twice.> DSSS CDMA uses the same principle to > spread the symbol over a large bandwidth by multiplying it with narrow > time pulses(chips).Not really. Spread spectrum is generally presented as holding the power constant and increasing the bandwidth through coding. The SNR at any one frequency goes _down_ as spreading goes up, it certainly doesn't stay constant.> Can we intutively say therefore, that when we de-modulate our spreaded > signal by remultiplying it with the PN sequene and then averaging it > over symbol period, we are in-effect using the statistical property > that noise over a large period(symbol period) will cancel out itself.Well, yes, but you can say that about any simple signal at a given symbol rate. Take white noise, feed it into an ideal spread-spectrum front end with gain = 1, and what comes out is white noise that's indistinguishable from what you had to start with. Take white noise plus a spread signal, feed it into that same front end (assume that synchronization has happened), and you get white noise plus the original, un-spread signal. So if you use an ideal system to spread, and then de-spread, the signal you don't change its noise properties, or error rates, or anything.> In other words the larger averaging interval we have(symbol period) > the more we lower noise, since noise is inherently random and > gaussian.In other words, spread spectrum does _nothing_ to the signal characteristics you're struggling with. Lowering the symbol rate increases the averaging interval, sure -- but that'll happen with or without spreading. What any of the multiple access systems do is to cleverly take a wide bandwidth channel (i.e. whatever the assignment is for that particular service) with high-power potential (because you've got great big base stations, and all those phones), and find ways to gracefully apportion that channel capacity amongst a whole bunch of base stations and phones. They all do this by making the signals to and from any phones in a given area mutually orthogonal. Spread spectrum is just one way among many, that has some strong advantages and some strong disadvantages. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●August 10, 20102010-08-10
ashu wrote:> Dear All, > I am trying to brush up my communication theory fundamentals. I have > simple(perhaps foolish!) question on intutive understanding of CDMA. > Shannon says that the capacity of a channel is directly proportional > to bandwidth of signal. In other others for a given SNR if increasing > B would give me more capacity.Yes.> DSSS CDMA uses the same principle to > spread the symbol over a large bandwidth by multiplying it with narrow > time pulses(chips).No. Just spreading a signal does not make for any gain in the Shannon theorem sense.> Can we intutively say therefore, that when we de-modulate our spreaded > signal by remultiplying it with the PN sequene and then averaging it > over symbol period, we are in-effect using the statistical property > that noise over a large period(symbol period) will cancel out itself.No. There is no difference if you are averaging over a spreading sequence or over a constant value per symbol.> In other words the larger averaging interval we have(symbol period) > the more we lower noise, since noise is inherently random and > gaussian. > > Can DSSS-CDMA be looked at from this perspective?Spread spectrum, CDMA and bandwidth expansion to gain the capacity are the different means to attain entirely different goals. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
Reply by ●August 10, 20102010-08-10
On Aug 10, 10:29=A0am, ashu <ashutosh.ghildi...@gmail.com> averred:> Shannon says that the capacity of a channel is directly proportional > to bandwidth of signal.I don't believe Shannon (or most of those who followed him) ever said that. The capacity of a channel is *not* ***directly proportional**** to the bandwidth of the signal; it is not even directly proportional to the bandwidth of the channel. The channel capacity increases with bandwidth but a law of diminishing returns sets in and the capacity approaches a finite limit (value depending on the ratio of signal power to noise density, which ratio is assumed to be fixed) as bandwidth goes to infinity. --Dilip Sarwate
Reply by ●August 10, 20102010-08-10
On 08/10/2010 11:33 AM, dvsarwate wrote:> On Aug 10, 10:29 am, ashu<ashutosh.ghildi...@gmail.com> averred: > >> Shannon says that the capacity of a channel is directly proportional >> to bandwidth of signal. > > I don't believe Shannon (or most of those who followed > him) ever said that. The capacity of a channel is *not* > ***directly proportional**** to the bandwidth of the signal; > it is not even directly proportional to the bandwidth of the > channel. The channel capacity increases with bandwidth > but a law of diminishing returns sets in and the capacity > approaches a finite limit (value depending on the ratio > of signal power to noise density, which ratio is assumed > to be fixed) as bandwidth goes to infinity.Actually the way he framed it was for a channel with a fixed SNR, in which case the channel capacity _would_ be proportional to bandwidth. But the "fixed SNR" channel isn't a very representative model, as I pointed out in my response. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●August 11, 20102010-08-11
On Aug 10, 10:30=A0pm, Tim Wescott <t...@seemywebsite.com> wrote:> On 08/10/2010 11:33 AM, dvsarwate wrote: > > > On Aug 10, 10:29 am, ashu<ashutosh.ghildi...@gmail.com> =A0averred: > > >> Shannon says that the capacity of a channel is directly proportional > >> to bandwidth of signal. > > > I don't believe Shannon (or most of those who followed > > him) ever said that. =A0The capacity of a channel is *not* > > ***directly proportional**** to the bandwidth of the signal; > > it is not even directly proportional to the bandwidth of the > > channel. =A0The channel capacity increases with bandwidth > > but a law of diminishing returns sets in and the capacity > > approaches a finite limit (value depending on the ratio > > of signal power to noise density, which ratio is assumed > > to be fixed) as bandwidth goes to infinity. > > Actually the way he framed it was for a channel with a fixed SNR, in > which case the channel capacity _would_ be proportional to bandwidth. > > But the "fixed SNR" channel isn't a very representative model, as I > pointed out in my response. > > -- > > Tim Wescott > Wescott Design Serviceshttp://www.wescottdesign.com > > Do you need to implement control loops in software? > "Applied Control Theory for Embedded Systems" was written for you. > See details athttp://www.wescottdesign.com/actfes/actfes.htmlI thank all of you for your interesting replies. Yes I meant fixed SNR and in that case Capaciy would be directly proprtional to Bandwidth. What I really want to understand is.. suppose my pulse period is fixed(hence pulse bandwidth), if the noise characterstics of channel are such that at the receiver, pulse is deeply burried in noise. In this kind of scenario, if I transmit multiple pulses, can I achieve reliable communication, by some sort of avergaging at the recevier ? Will transmitting more pulses help ? regards and thanks ashu
Reply by ●August 11, 20102010-08-11
On Aug 10, 3:30=A0pm, Tim Wescott <t...@seemywebsite.com> wrote:> > Actually the way he framed it was for a channel with a fixed SNR, in > which case the channel capacity _would_ be proportional to bandwidth. > > But the "fixed SNR" channel isn't a very representative model, as I > pointed out in my response. >What the OP wrote was "Shannon says that the capacity of a channel is directly proportional to bandwidth of signal." without any mention of SNR, and that was what I was complaining about. While the OP mentioned SNR in the next sentence, as was pointed out in a different very recent thread with title "Tangled up with Shaonnon Bound", there are different definitions of what SNR means that are used in communication system studies. Thus, unless it is made clear what is meant by SNR, the OP's assertion "In other others for a given SNR if increasing B would give me more capacity." is not necessarily true. Even if we give credence to the OP's latest assertion that he was thinking of fixed SNR (presumably meaning the quantity P/N_0W in the capacity formula C =3D W log(1 + P/N_0W) being fixed as W varies, so that P is increasing with W and thus C is proportional to W), all along, the very next sentence in the original post (referring to direct-sequence spread-spectrum) suggests that the OP really was thinking of fixed SNR as meaning Eb/N0 being fixed. As both Tim and Vladimir pointed out, direct-sequence modulation does not change Eb/N0 and so the same BER performance is obtained regardless of whether the signal is spread or not. Moral: you can get all kinds of wonderful results by defining SNR appropriately, but you can't fake the BEND* ratio Eb/N0, and that's what really counts. --Dilip Sarwate *BEND =3D bit-energy-to-noise-density
Reply by ●August 11, 20102010-08-11
On 08/11/2010 01:11 AM, ashu wrote:> On Aug 10, 10:30 pm, Tim Wescott<t...@seemywebsite.com> wrote: >> On 08/10/2010 11:33 AM, dvsarwate wrote: >> >>> On Aug 10, 10:29 am, ashu<ashutosh.ghildi...@gmail.com> averred: >> >>>> Shannon says that the capacity of a channel is directly proportional >>>> to bandwidth of signal. >> >>> I don't believe Shannon (or most of those who followed >>> him) ever said that. The capacity of a channel is *not* >>> ***directly proportional**** to the bandwidth of the signal; >>> it is not even directly proportional to the bandwidth of the >>> channel. The channel capacity increases with bandwidth >>> but a law of diminishing returns sets in and the capacity >>> approaches a finite limit (value depending on the ratio >>> of signal power to noise density, which ratio is assumed >>> to be fixed) as bandwidth goes to infinity. >> >> Actually the way he framed it was for a channel with a fixed SNR, in >> which case the channel capacity _would_ be proportional to bandwidth. >> >> But the "fixed SNR" channel isn't a very representative model, as I >> pointed out in my response. >> >> -- >> >> Tim Wescott >> Wescott Design Serviceshttp://www.wescottdesign.com >> >> Do you need to implement control loops in software? >> "Applied Control Theory for Embedded Systems" was written for you. >> See details athttp://www.wescottdesign.com/actfes/actfes.html > > I thank all of you for your interesting replies. > > Yes I meant fixed SNR and in that case Capaciy would be directly > proprtional to Bandwidth. > > What I really want to understand is.. suppose my pulse period is > fixed(hence pulse bandwidth), if the noise characterstics of channel > are such that at the receiver, pulse is deeply burried in noise. In > this kind of scenario, if I transmit multiple pulses, can I achieve > reliable communication, by some sort of avergaging at the recevier ?Yes. In a sense this is what you are doing when you increase your symbol length (consider a 1ms pulse as 'just' 1000 1us pulses). It is also what you are doing when you send long symbols over a spread-spectrum link, although you then have synchronization to deal with.> Will transmitting more pulses help ?Yes, but that may not be the best solution, or it may only be a small part of the best solution. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●August 11, 20102010-08-11
> > Yes. =A0In a sense this is what you are doing when you increase your > symbol length (consider a 1ms pulse as 'just' 1000 1us pulses). =A0It is > also what you are doing when you send long symbols over a > spread-spectrum link, although you then have synchronization to deal with=.> > > Will transmitting more pulses help ? > > Yes, but that may not be the best solution, or it may only be a small > part of the best solution. > > -- > > Tim Wescott > Wescott Design Serviceshttp://www.wescottdesign.com > > Do you need to implement control loops in software? > "Applied Control Theory for Embedded Systems" was written for you. > See details athttp://www.wescottdesign.com/actfes/actfes.html- Hide quote=d text -> > - Show quoted text -I think I'm getting an insight here... repeating the same symbol 1000 times and averaging the result to obtain 1 bit of info would be analagous to reducing the video BW on a spectrum analyzer i.e. it reduces the POST DETECTION BW, and while this does help it would be (much) more effective to reduce the resolution BW i.e. reduce the PRE DETECTION BW. To do this you must reduce the symbol rate. So instead of sending 1000 symbols that carry 1 bit, it is better to send 1 symbol that is 1000 time longer and hence requires 1000 less pre dection BW... is that it? Mark
Reply by ●August 11, 20102010-08-11
On 08/11/2010 10:07 AM, Mark wrote:> >> >> Yes. In a sense this is what you are doing when you increase your >> symbol length (consider a 1ms pulse as 'just' 1000 1us pulses). It is >> also what you are doing when you send long symbols over a >> spread-spectrum link, although you then have synchronization to deal with. >> >>> Will transmitting more pulses help ? >> >> Yes, but that may not be the best solution, or it may only be a small >> part of the best solution. >> >> -- >> >> Tim Wescott >> Wescott Design Serviceshttp://www.wescottdesign.com >> >> Do you need to implement control loops in software? >> "Applied Control Theory for Embedded Systems" was written for you. >> See details athttp://www.wescottdesign.com/actfes/actfes.html- Hide quoted text - >> >> - Show quoted text - > > I think I'm getting an insight here... > > repeating the same symbol 1000 times and averaging the result to > obtain 1 bit of info would be analagous to reducing the video BW on a > spectrum analyzer i.e. it reduces the POST DETECTION BW, and while > this does help it would be (much) more effective to reduce the > resolution BW i.e. reduce the PRE DETECTION BW. To do this you must > reduce the symbol rate. > > So instead of sending 1000 symbols that carry 1 bit, it is better to > send 1 symbol that is 1000 time longer and hence requires 1000 less > pre dection BW... is that it?Well... If you knew the transmitter were repeating each bit 1000 times, you wouldn't detect each bit and average after the fact -- you'd just average over those 1000 'bits' as one symbol. This is your "pre detection" bandwidth reduction. Actually doing a hard detection on each bit, then doing a majority vote on the result is going to have worse performance than averaging the whole 1000 bits as a lump, but better than any individual bit detection. How much better depends on Eb/No. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
